Which method is faster for replacing the last occurrence of a substring in a string?

Question

I have a string which is a combination of "S"s and "C"s. There is at least one occurrence of each.

I want to find the last occurrence of "CS" and change it to "SC".

I have two methods (so far):

P = P[::-1].replace("SC", "CS", 1)[::-1]

and

P = P[:P.rfind("CS")] + "SC" + P[P.rfind("CS") + 2:]

Which line is likely to be faster? Also, is there a quicker way to achieve what I'm doing?

You can test this by running both functions times and measuring the time difference between start and end using time.time(). — Nathan, Apr 07 '18 at 14:41
You should look into the `timeit` module. If you are working in the iPython console, look into the `%timeit` magic command. You can answer this kind of question ("which line is likely to be faster?") on your own. — Rory Daulton, Apr 07 '18 at 14:47

Austin · Accepted Answer · 2018-04-07T16:08:48.090

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This is one way:

>>> s = 'CSsomethingSCSagainCSsomething'
>>> 'SC'.join(s.rsplit('CS', 1))
CSsomethingSCSagainSCsomething

Syntax:

new_substring.join(str.rsplit(old_substring, occurance))

edited Apr 07 '18 at 16:08

answered Apr 07 '18 at 14:48

Austin

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it beats OP attempt in my bench. – Jean-François Fabre Apr 07 '18 at 14:49
@Jean-FrançoisFabre Also, so this will only replace the last occurrence, right? – Adi219 Apr 07 '18 at 14:50
@Adi219 but I benched my answer afterwards, and it appears that it was slower... yes, it replaces the last occurrence. – Jean-François Fabre Apr 07 '18 at 14:50
@Adi219 Yes, it replaces the last occurrence only. – Austin Apr 07 '18 at 14:52
Thanks! This helps a lot :) – Adi219 Apr 07 '18 at 14:52
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@Adi219 Probably you may wait for other answers and accept only if you find this one good. – Austin Apr 07 '18 at 14:54
@theausome I doubt python can do it any faster. Let's see. – Jean-François Fabre Apr 07 '18 at 14:54
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@Adi219 you should accept it. You can accept another one if a better one arises (but I doubt it!) – Jean-François Fabre Apr 07 '18 at 14:55
@Jean-FrançoisFabre BTW, I'm doing Google Code Jam today and I want to make sure my code is as optimised as possible, so this will help a lot! :) – Adi219 Apr 07 '18 at 14:55
If you'll be doing this multiple times, you can save ~12% by caching the lookup of `'SC'.join`. `x = 'SC'.join; x(s.rsplit('CS', 1))`. (You can't apply the same trick to `s.rsplit` because, presumably, `s` would be different every time you use this.) – chepner Apr 07 '18 at 15:17
(And using `methodcaller('rsplit', 'CS', 1)(s)` is slightly slower than just using `s.rsplit('CS', 1)`.) – chepner Apr 07 '18 at 15:18
@chepner You're right in that I'm going to repeat this operation many times (and with different strings each time). Please could you elaborate on (both of) your points? (As I don't fully understand) – Adi219 Apr 07 '18 at 15:59
This example code, as shown, doesn't run as it has unbalanced parentheses. – cdlane Apr 07 '18 at 16:07
@cdlane edited now. That was a mistake I made on my second edit. Thanks. – Austin Apr 07 '18 at 16:09
@Jean-FrançoisFabre There seems to be an answer which claims to do it about 20% faster – Adi219 Apr 07 '18 at 16:21

score 2 · Answer 2 · answered Apr 07 '18 at 16:06

2

In my crude timing, this beats the rsplit() and join() solution by 20%:

head, _, tail = string.rpartition('CS')
new_string = f"{head}SC{tail}"

I does depend on Python 3.6+, of course.

answered Apr 07 '18 at 16:06

cdlane

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yes it's faster. But with a classical `format(head=head,...)` it doesn't work as fast. Python 3.6 only. Nice – Jean-François Fabre Apr 07 '18 at 20:05

Which method is faster for replacing the last occurrence of a substring in a string?

2 Answers2